##Introduction
When you are asked to find k so that the following function is continuous, the task often feels like a puzzle where the missing piece is a single number that guarantees smoothness. Continuity is a fundamental property in calculus that ensures a function’s graph can be drawn without lifting the pen. In practical terms, this means that at every point in its domain, the limit of the function as you approach the point from the left must equal the limit from the right, and both must equal the actual function value at that point Turns out it matters..
This is the bit that actually matters in practice It's one of those things that adds up..
In this article we will work through a concrete example, step by step, to illustrate the process. The function we will consider is
[ f(x)= \begin{cases} \dfrac{x^{2}-9}{x-3}, & x\neq 3\[6pt] k, & x=3 \end{cases} ]
At first glance the expression (\frac{x^{2}-9}{x-3}) looks problematic because it has a denominator that becomes zero at (x=3). That said, algebraic simplification reveals that the fraction can be reduced, and the limit as (x) approaches 3 exists. Our goal is to determine the value of k that makes the function continuous at the critical point (x=3) Surprisingly effective..
The official docs gloss over this. That's a mistake.
By the end of this guide you will not only have the answer (which turns out to be k = 6) but also a clear, repeatable method you can apply to any piecewise function you encounter.
Steps
To find k so that the function is continuous, follow these logical steps. Each step is presented as a bullet list for easy reference.
-
Identify the point of potential discontinuity
Look for the value of (x) where the definition of the function changes. In our example, the only place where the rule switches is at (x=3). -
Compute the left‑hand limit
Evaluate (\displaystyle \lim_{x\to 3^-} f(x)). Since for (x\neq 3) the function simplifies to (x+3), the left‑hand limit is (3+3 = 6) Nothing fancy.. -
Compute the right‑hand limit
Evaluate (\displaystyle \lim_{x\to 3^+} f(x)). The same simplification applies, so the right‑hand limit is also (6). -
Set the limits equal to the function value at the point
For continuity we need
[ \lim_{x\to 3^-} f(x)=\lim_{
Setting the left‑hand limit, the right‑hand limit, and the defined value equal gives
[ \lim_{x\to 3^-} f(x)=\lim_{x\to 3^+} f(x)=f(3)=k. ]
Since the simplified expression (\frac{x^{2}-9}{x-3}=x+3) yields a limit of (6) from both sides, we obtain
[ k = 6. ]
Thus the piecewise definition becomes
[ f(x)=\begin{cases} \dfrac{x^{2}-9}{x-3}, & x\neq 3,\[6pt] 6, & x=3, \end{cases} ]
and the graph can be drawn without lifting the pen at (x=3); the function is now continuous everywhere Nothing fancy..
General approach
Whenever a piecewise function contains a parameter that appears only at a point of definition change, the same procedure applies:
- Locate the point where the formula switches.
- Simplify the expression on the domain where it is defined (often by factoring or canceling common factors).
- Compute the left‑hand and right‑hand limits as the variable approaches the point.
- Equate those limits to the value assigned at the point (the unknown parameter).
- Solve the resulting equation for the parameter.
If
